跳转至

83.删除排序链表中的重复元素

对于重复的只取最后一个ListNode即可，通过递归依次去重

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null){
            return null; //终止递归
        }
        int val = head.val;
        while(head.next != null && val == head.next.val){
            head = head.next;
        }
        head.next = deleteDuplicates(head.next);
        return head; 
    }
}

另外一种解法，取重复的第一个为current，然后如果是重复的就跳过。

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
//1->1->2->3->3
public ListNode deleteDuplicates(ListNode head) {
    ListNode current = head;
    while (current != null && current.next != null) {
        if (current.next.val == current.val) {
            //跳过
            current.next = current.next.next;
        } else {
            current = current.next;
        }
    }
    return head;
}